Question

For each of the following, decide if the given vector field is a gradient of a function f. If so, give the function f nd enter it as your answer; if not, enter the word NONE for your answer and be sure that you are able to explain why this is the case.

(b) 6xcos(y^2)î + (2ycos((x^2) + (y^2)))j

(c) [6cos((x^2) + (2y^2)) - 12(x^2)sin((x^2) + (2y^2))] - [24xysin((x^2) + (2y^2))]

Answer 1

@mathmale
HELP again??

Answer 2

@Disco619 this is Calc 3 I hope you can do it!

Answer 3

@TuringTest

Answer 4

@Snowfire Can you help me with this question?

Answer 5

It's either @Snowfire or @Denu8 . Denu is offline. Snow might be able to do this... i hope.

Answer 6

@Isaiah.Feynman

Answer 7

if a vector field is conservative, it is the gradient of some potential function
so we need to know if these fields are conservative

Answer 8

@ranga I THINK YOU CAN DO THIS!

Answer 9

a vector field \[\vec F=P\hat i+Q\hat j\]is conservative if\[{\partial P\over \partial y}={\partial Q\over\partial x}\]

Answer 10

take those partials and compare them

if they are equal, the vector field is the gradient of some potential function

Answer 11

as far as finding the function, that takes a bit of integration

the process is outlined nicely here in example 2:
http://tutorial.math.lamar.edu/Classes/CalcIII/ConservativeVectorField.aspx

Answer 12

Keep going @TuryingTest I still need help lol... I can't seem to get the partial of either of them. and what do i do AFTER i know they are conservative?

Answer 13

if you can't get the partial then there's not much point in trying to fund a possibly non-existent potential function, but you would take note of the fact that by definition\[\vec F=\nabla f=\frac{\partial f}{\partial x}\hat i+\frac{\partial f}{\partial y}\hat j\implies P=\frac{\partial f}{\partial x}, ~~Q=\frac{\partial f}{\partial y}\]

Answer 14

for the first one, 6xcos(y^2)î + (2ycos((x^2) + (y^2)))j
P= 6xcos(y^2)
what the partial of that wrt y?

Answer 15

6xsin(y^2)(2)...?

Answer 16

dropped a y

Answer 17

Oh 6x(sin(y^2))(2y) right?

Answer 18

right, not what's partial wrt x of Q= (2ycos((x^2) + (y^2)))

Answer 19

now*

Answer 20

2ycos(x^2)(2x) + (y^2)x ?

Answer 21

erm, not if I'm reading those parentheses right
is it\[Q=2y\cos(x^2+y^2)\]?

Answer 22

yes that is what Q is

Answer 23

so then treating y as constant...\[\frac{\partial Q}{\partial x}=-2y\sin(x^2+y^2)\frac{\partial }{\partial x}(x^2+y^2)=-2xy\sin(x^2+y^2)\]

Answer 24

oh ok i think i was still doing wrt y like we did for P. OK so now what
and seriously thanks so much for working with me here

Answer 25

oh typo, should be a 4 at the end\[\frac{\partial Q}{\partial x}=-2y\sin(x^2+y^2)\frac{\partial }{\partial x}(x^2+y^2)=-4xy\sin(x^2+y^2)\]

Answer 26

you're welcome
anyway, is \[{\partial P\over \partial y}={\partial Q\over\partial x}\]?

Answer 27

umm no?

Answer 28

so then \(\vec F\) is not conservative, which means it has no potential function \(f\), so we're done

Answer 29

I'll let you check if\[{\partial P\over \partial y}={\partial Q\over\partial x}\]for the next oneon your own

Answer 30

@mathmale I need help with part C

Answer 31

@TuringTest This is the work I have so far for part C.
@mathmale could ya help me out?

Answer 32

@ranga Can you help me on this?

Answer 33

I NEED THIS QUESTION ANSWERED BY 5:00 TONIGHT!